How do you find the derivative of ${cos}^{2} (x^{2} - 3 x)$ $cos^2 (x^2 - 3x)$ ?

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Answer 1 · 2016-06-27T04:38:20

$(6 - 4 x) cos (x^{2} - 3 x) sin (x^{2} - 3 x)$ $(6-4x) cos (x^2-3x) sin (x^2 -3x)$

First differentiate cosine function w.r.t $x^{2} - 3 x$ $x^2-3x$ and then differentiate $x^{2} - 3 x$ $x^2-3x$ w.r.t x.............. Chain rule

Derivative would be $- 2 cos (x^{2} - 3 x) sin (x^{2} - 3 x) (2 x - 3)$ $-2cos(x^2 -3x) sin(x^2-3x) (2x-3)$

$(6 - 4 x) cos (x^{2} - 3 x) sin (x^{2} - 3 x)$ $(6-4x) cos (x^2-3x) sin (x^2 -3x)$

How do you find the derivative of cos^2 (x^2 - 3x)cos2(x2−3x)cos^2 (x^2 - 3x)?